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In \( \triangle \mathrm{ABC} \) and \( \triangle \mathrm{DEF}, \mathrm{AB}=\mathrm{DE}, \mathrm{AB} \| \mathrm{DE}, \mathrm{BC}=\mathrm{EF} \) and \( \mathrm{BC} \| EF \). Vertices \( \mathrm{A}, \mathrm{B} \) and \( \mathrm{C} \) are joined to vertices D, E and F respectively (see below figure).
Show that
(i) quadrilateral ABED is a parallelogram
(ii) quadrilateral \( \mathrm{BEFC} \) is a parallelogram
(iii) \( \mathrm{AD} \| \mathrm{CF} \) and \( \mathrm{AD}=\mathrm{CF} \)
(iv) quadrilateral ACFD is a parallelogram
(v) \( \mathrm{AC}=\mathrm{DF} \)
(vi) \( \triangle \mathrm{ABC} \equiv \triangle \mathrm{DEF} \).
Given:
In \( \triangle \mathrm{ABC} \) and \( \triangle \mathrm{DEF}, \mathrm{AB}=\mathrm{DE}, \mathrm{AB} \| \mathrm{DE}, \mathrm{BC}=\mathrm{EF} \) and \( \mathrm{BC} \| EF \).
To do :
We have to show that
(i) quadrilateral ABED is a parallelogram
(ii) quadrilateral \( \mathrm{BEFC} \) is a parallelogram
(iii) \( \mathrm{AD} \| \mathrm{CF} \) and \( \mathrm{AD}=\mathrm{CF} \)
(iv) quadrilateral ACFD is a parallelogram
(v) \( \mathrm{AC}=\mathrm{DF} \)
(vi) \( \triangle \mathrm{ABC} \equiv \triangle \mathrm{DEF} \).
Solution :
(i) In quadrilateral $ABED$,
$AB=DE$
$AB \| DE$
We know that,
A parallelogram is a quadrilateral with opposite sides equal and parallel to each other.
Therefore,
$ABED$ is a parallelogram.
(ii) In quadrilateral $BECF$,
$BC=EF$
$BC \| EF$
Therefore,
$BECF$ is a parallelogram.
(iii) $ABED$ and $BEFC$ are parallelograms.
We know that,
Opposite sides of a parallelogram are equal and parallel.
This implies,
$AD = BE$
$BE = CF$
$\Rightarrow AD = CF$
$AD \| BE$
$BE \| CF$
$\Rightarrow AD \| CF$
(iv) In quadrilateral $ACFD$,
$AD=CF$ (Proved)
$AD \| CF$ (Proved)
Therefore,
$ACFD$ is a parallelogram.
(v) $ACFD$ is a parallelogram
This implies,
$AC \| DF$
$AC = DF$
(vi) In $\triangle ABC$ and $\triangle DEF$,
$AB = DE$ (Given)
$BC = EF$ (Given)
$AC = DF$ (Opposite sides of a parallelogram are equal)
Therefore, by SSS congruency, we get,
$\triangle ABC \cong \triangle DEF$.